Integrand size = 112, antiderivative size = 29 \begin {dmath*} \int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \left (1+e^{2 e^{1-5 e^{1-x}+5 x}} x^2-2 \log (x)+e^{-5 e^{1-x}+5 x} \left (-10 e x-10 e^{2-x} x\right ) \log (x)\right ) \, dx=x^{1+\frac {e^{-2 e^{1+5 \left (-e^{1-x}+x\right )}}}{x^2}} \end {dmath*}
Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \begin {dmath*} \int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \left (1+e^{2 e^{1-5 e^{1-x}+5 x}} x^2-2 \log (x)+e^{-5 e^{1-x}+5 x} \left (-10 e x-10 e^{2-x} x\right ) \log (x)\right ) \, dx=x^{1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \end {dmath*}
Integrate[(x^(-2 + 1/(E^(2*E^(1 - 5*E^(1 - x) + 5*x))*x^2))*(1 + E^(2*E^(1 - 5*E^(1 - x) + 5*x))*x^2 - 2*Log[x] + E^(-5*E^(1 - x) + 5*x)*(-10*E*x - 10*E^(2 - x)*x)*Log[x]))/E^(2*E^(1 - 5*E^(1 - x) + 5*x)),x]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int e^{-2 e^{5 x-5 e^{1-x}+1}} x^{\frac {e^{-2 e^{5 x-5 e^{1-x}+1}}}{x^2}-2} \left (e^{2 e^{5 x-5 e^{1-x}+1}} x^2+e^{5 x-5 e^{1-x}} \left (-10 e^{2-x} x-10 e x\right ) \log (x)-2 \log (x)+1\right ) \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (e^{-2 e^{5 x-5 e^{1-x}+1}} x^{\frac {e^{-2 e^{5 x-5 e^{1-x}+1}}}{x^2}-2}-2 e^{-2 e^{5 x-5 e^{1-x}+1}} x^{\frac {e^{-2 e^{5 x-5 e^{1-x}+1}}}{x^2}-2} \log (x)-10 e^{4 x-5 e^{1-x}-2 e^{5 x-5 e^{1-x}+1}+1} \left (e^x+e\right ) x^{\frac {e^{-2 e^{5 x-5 e^{1-x}+1}}}{x^2}-1} \log (x)+x^{\frac {e^{-2 e^{5 x-5 e^{1-x}+1}}}{x^2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int e^{-2 e^{5 x-5 e^{1-x}+1}} x^{\frac {e^{-2 e^{5 x-5 e^{1-x}+1}}}{x^2}-2}dx+2 \int \frac {\int e^{-2 e^{5 x-5 e^{1-x}+1}} x^{\frac {e^{-2 e^{5 x-5 e^{1-x}+1}}}{x^2}-2}dx}{x}dx+10 \int \frac {\int e^{4 x-5 e^{1-x}-2 e^{5 x-5 e^{1-x}+1}+2} x^{\frac {e^{-2 e^{5 x-5 e^{1-x}+1}}}{x^2}-1}dx}{x}dx+10 \int \frac {\int e^{5 x-5 e^{1-x}-2 e^{5 x-5 e^{1-x}+1}+1} x^{\frac {e^{-2 e^{5 x-5 e^{1-x}+1}}}{x^2}-1}dx}{x}dx-2 \log (x) \int e^{-2 e^{5 x-5 e^{1-x}+1}} x^{\frac {e^{-2 e^{5 x-5 e^{1-x}+1}}}{x^2}-2}dx-10 \log (x) \int e^{4 x-5 e^{1-x}-2 e^{5 x-5 e^{1-x}+1}+2} x^{\frac {e^{-2 e^{5 x-5 e^{1-x}+1}}}{x^2}-1}dx-10 \log (x) \int e^{5 x-5 e^{1-x}-2 e^{5 x-5 e^{1-x}+1}+1} x^{\frac {e^{-2 e^{5 x-5 e^{1-x}+1}}}{x^2}-1}dx+\int x^{\frac {e^{-2 e^{5 x-5 e^{1-x}+1}}}{x^2}}dx\) |
Int[(x^(-2 + 1/(E^(2*E^(1 - 5*E^(1 - x) + 5*x))*x^2))*(1 + E^(2*E^(1 - 5*E ^(1 - x) + 5*x))*x^2 - 2*Log[x] + E^(-5*E^(1 - x) + 5*x)*(-10*E*x - 10*E^( 2 - x)*x)*Log[x]))/E^(2*E^(1 - 5*E^(1 - x) + 5*x)),x]
3.14.28.3.1 Defintions of rubi rules used
Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90
\[x^{\frac {{\mathrm e}^{-2 \,{\mathrm e}^{1-5 \,{\mathrm e}^{1-x}+5 x}}}{x^{2}}} x\]
int((x^2*exp(exp(1)*exp(-5*exp(1-x)+5*x))^2+(-10*x*exp(1)*exp(1-x)-10*x*ex p(1))*ln(x)*exp(-5*exp(1-x)+5*x)-2*ln(x)+1)*exp(ln(x)/x^2/exp(exp(1)*exp(- 5*exp(1-x)+5*x))^2)/x^2/exp(exp(1)*exp(-5*exp(1-x)+5*x))^2,x)
Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \begin {dmath*} \int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \left (1+e^{2 e^{1-5 e^{1-x}+5 x}} x^2-2 \log (x)+e^{-5 e^{1-x}+5 x} \left (-10 e x-10 e^{2-x} x\right ) \log (x)\right ) \, dx=x x^{\frac {e^{\left (-2 \, e^{\left ({\left ({\left (5 \, x + 1\right )} e - 5 \, e^{\left (-x + 2\right )}\right )} e^{\left (-1\right )}\right )}\right )}}{x^{2}}} \end {dmath*}
integrate((x^2*exp(exp(1)*exp(-5*exp(1-x)+5*x))^2+(-10*x*exp(1)*exp(1-x)-1 0*x*exp(1))*log(x)*exp(-5*exp(1-x)+5*x)-2*log(x)+1)*exp(log(x)/x^2/exp(exp (1)*exp(-5*exp(1-x)+5*x))^2)/x^2/exp(exp(1)*exp(-5*exp(1-x)+5*x))^2,x, alg orithm=\
Timed out. \begin {dmath*} \int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \left (1+e^{2 e^{1-5 e^{1-x}+5 x}} x^2-2 \log (x)+e^{-5 e^{1-x}+5 x} \left (-10 e x-10 e^{2-x} x\right ) \log (x)\right ) \, dx=\text {Timed out} \end {dmath*}
integrate((x**2*exp(exp(1)*exp(-5*exp(1-x)+5*x))**2+(-10*x*exp(1)*exp(1-x) -10*x*exp(1))*ln(x)*exp(-5*exp(1-x)+5*x)-2*ln(x)+1)*exp(ln(x)/x**2/exp(exp (1)*exp(-5*exp(1-x)+5*x))**2)/x**2/exp(exp(1)*exp(-5*exp(1-x)+5*x))**2,x)
\begin {dmath*} \int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \left (1+e^{2 e^{1-5 e^{1-x}+5 x}} x^2-2 \log (x)+e^{-5 e^{1-x}+5 x} \left (-10 e x-10 e^{2-x} x\right ) \log (x)\right ) \, dx=\int { \frac {{\left (x^{2} e^{\left (2 \, e^{\left (5 \, x - 5 \, e^{\left (-x + 1\right )} + 1\right )}\right )} - 10 \, {\left (x e + x e^{\left (-x + 2\right )}\right )} e^{\left (5 \, x - 5 \, e^{\left (-x + 1\right )}\right )} \log \left (x\right ) - 2 \, \log \left (x\right ) + 1\right )} x^{\frac {e^{\left (-2 \, e^{\left (5 \, x - 5 \, e^{\left (-x + 1\right )} + 1\right )}\right )}}{x^{2}}} e^{\left (-2 \, e^{\left (5 \, x - 5 \, e^{\left (-x + 1\right )} + 1\right )}\right )}}{x^{2}} \,d x } \end {dmath*}
integrate((x^2*exp(exp(1)*exp(-5*exp(1-x)+5*x))^2+(-10*x*exp(1)*exp(1-x)-1 0*x*exp(1))*log(x)*exp(-5*exp(1-x)+5*x)-2*log(x)+1)*exp(log(x)/x^2/exp(exp (1)*exp(-5*exp(1-x)+5*x))^2)/x^2/exp(exp(1)*exp(-5*exp(1-x)+5*x))^2,x, alg orithm=\
integrate((x^2*e^(2*e^(5*x - 5*e^(-x + 1) + 1)) - 10*(x*e + x*e^(-x + 2))* e^(5*x - 5*e^(-x + 1))*log(x) - 2*log(x) + 1)*x^(e^(-2*e^(5*x - 5*e^(-x + 1) + 1))/x^2 - 2)*e^(-2*e^(5*x - 5*e^(-x + 1) + 1)), x)
\begin {dmath*} \int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \left (1+e^{2 e^{1-5 e^{1-x}+5 x}} x^2-2 \log (x)+e^{-5 e^{1-x}+5 x} \left (-10 e x-10 e^{2-x} x\right ) \log (x)\right ) \, dx=\int { \frac {{\left (x^{2} e^{\left (2 \, e^{\left (5 \, x - 5 \, e^{\left (-x + 1\right )} + 1\right )}\right )} - 10 \, {\left (x e + x e^{\left (-x + 2\right )}\right )} e^{\left (5 \, x - 5 \, e^{\left (-x + 1\right )}\right )} \log \left (x\right ) - 2 \, \log \left (x\right ) + 1\right )} x^{\frac {e^{\left (-2 \, e^{\left (5 \, x - 5 \, e^{\left (-x + 1\right )} + 1\right )}\right )}}{x^{2}}} e^{\left (-2 \, e^{\left (5 \, x - 5 \, e^{\left (-x + 1\right )} + 1\right )}\right )}}{x^{2}} \,d x } \end {dmath*}
integrate((x^2*exp(exp(1)*exp(-5*exp(1-x)+5*x))^2+(-10*x*exp(1)*exp(1-x)-1 0*x*exp(1))*log(x)*exp(-5*exp(1-x)+5*x)-2*log(x)+1)*exp(log(x)/x^2/exp(exp (1)*exp(-5*exp(1-x)+5*x))^2)/x^2/exp(exp(1)*exp(-5*exp(1-x)+5*x))^2,x, alg orithm=\
integrate((x^2*e^(2*e^(5*x - 5*e^(-x + 1) + 1)) - 10*(x*e + x*e^(-x + 2))* e^(5*x - 5*e^(-x + 1))*log(x) - 2*log(x) + 1)*x^(e^(-2*e^(5*x - 5*e^(-x + 1) + 1))/x^2)*e^(-2*e^(5*x - 5*e^(-x + 1) + 1))/x^2, x)
Timed out. \begin {dmath*} \int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \left (1+e^{2 e^{1-5 e^{1-x}+5 x}} x^2-2 \log (x)+e^{-5 e^{1-x}+5 x} \left (-10 e x-10 e^{2-x} x\right ) \log (x)\right ) \, dx=\int -\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{-2\,\mathrm {e}\,{\mathrm {e}}^{5\,x-5\,{\mathrm {e}}^{1-x}}}\,\ln \left (x\right )}{x^2}}\,{\mathrm {e}}^{-2\,\mathrm {e}\,{\mathrm {e}}^{5\,x-5\,{\mathrm {e}}^{1-x}}}\,\left (2\,\ln \left (x\right )-x^2\,{\mathrm {e}}^{2\,\mathrm {e}\,{\mathrm {e}}^{5\,x-5\,{\mathrm {e}}^{1-x}}}+{\mathrm {e}}^{5\,x-5\,{\mathrm {e}}^{1-x}}\,\ln \left (x\right )\,\left (10\,x\,\mathrm {e}+10\,x\,\mathrm {e}\,{\mathrm {e}}^{1-x}\right )-1\right )}{x^2} \,d x \end {dmath*}
int(-(exp((exp(-2*exp(1)*exp(5*x - 5*exp(1 - x)))*log(x))/x^2)*exp(-2*exp( 1)*exp(5*x - 5*exp(1 - x)))*(2*log(x) - x^2*exp(2*exp(1)*exp(5*x - 5*exp(1 - x))) + exp(5*x - 5*exp(1 - x))*log(x)*(10*x*exp(1) + 10*x*exp(1)*exp(1 - x)) - 1))/x^2,x)